lua call function from a string with function name

Question

Is it possible in lua to execute a function from a string representing its name?
i.e: I have the string x = "foo", is it possible to do x() ?

If yes what is the syntax ?

Answer 1

Names are not unique, there can be many functions names foo in different namespaces. But _G['foo'] is foo in the global namespace.

Answer 2

It sounds like you want to do an 'eval', which is supported in Lua like so:

assert(loadstring(x))()

You'll probably want to concatenate the "()" onto x first, though.

Answer 3

loadstring is not the answer here. For starters you would need a return in the string, and other details I won't go into.

THC4k has the right idea; if you have the function name in the variable x, then the call you want is

_G[x](arg1, arg2, ...)

Answer 4

To call a function in the global namespace (as mentioned by @THC4k) is easily done, and does not require loadstring().

x='foo'
_G[x]() -- calls foo from the global namespace

You would need to use loadstring() (or walk each table) if the function in another table, such as if x='math.sqrt'.

If loadstring() is used you would want to not only append parenthesis with ellipse (...) to allow for parameters, but also add return to the front.

x='math.sqrt'
print(assert(loadstring('return '..x..'(...)'))(25)) --> 5

or walk the tables:

function findfunction(x)
  assert(type(x) == "string")
  local f=_G
  for v in x:gmatch("[^%.]+") do
    if type(f) ~= "table" then
       return nil, "looking for '"..v.."' expected table, not "..type(f)
    end
    f=f[v]
  end
  if type(f) == "function" then
    return f
  else
    return nil, "expected function, not "..type(f)
  end
end

x='math.sqrt'
print(assert(findfunction(x))(121)) -->11